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\title{\heiti\zihao{2} 习题2.6}
\author{中书君}
\date{\songti 2021年1月13日}

\begin{document}
\maketitle
\section{计算下列数列上下极限:}
\subsection{$a_{n}=\sqrt[n]{1+2^{n(-1)^{n}}}$}
\textbf{解}\quad
$\lim _{k \rightarrow \infty} a_{2 k}=\lim _{n \rightarrow \infty} \sqrt[ 2 k]{1+2^{2 k}}=2,\quad \lim _{k \rightarrow \infty} a_{2 k+1}=\lim _{n \rightarrow \infty}  \sqrt[2 k]{1+\frac{1}{2^{2 k}}}=1$
\par 故$\varlimsup \{a_{n}\}=2, \varliminf \{a_{n}\}=1$


\subsection{$a_{n}=1+\frac{n \cdot \cos \left(\frac{n \pi}{2}\right)}{n+1}$}
\textbf{解}\quad
$x_{1}=1, x_{2}=1-\frac{2}{3}, x_{3}=1, x_{4}=1+\frac{4}{5}, \cdots$

$\inf \left\{x_{n}\right\}=0 ; \quad \sup \left\{x_{n}\right\}=2 ; \quad \varliminf _{n \rightarrow \infty} x_{n}=0 ; \quad \varlimsup_{n \rightarrow \infty} x_{n}=2$

\subsection{$a_{n}=\frac{n}{3}-\left[\frac{n}{3}\right]$}
\textbf{解}\quad
$a_{3 k}=0,a_{3k+1} =\frac{1}{3},a_{3 k+2}=\frac{2}{3}$

$\therefore \varliminf _{n \rightarrow \infty} \{a_{n}\}=0 \quad \varlimsup _{n \rightarrow \infty} \{a_{n}\}=\frac{2}{3}$

\subsection{$a_{n+1}=\left\{\begin{array}{ll}\frac{a_{n}}{2}, & n \text { 是偶数, } \\ \frac{1+a_{n}}{2}, & n \text { 是奇数. }\end{array}\right.$}
\textbf{解}\quad
$$
	\begin{aligned}
		& a_{2 k+2}=\frac{1+a_{2k+ 1}}{2}=\frac{1+\frac{a_{2 k}}{2}}{2}=\frac{2+a_{2 k}}{4} \\ &a_{2 k+1} =\frac{a_{2 k}}{2}=\frac{1+a_{2 k-1}}{4} \end{aligned}
$$

$\lim _{k \rightarrow \infty} a_{2 k+2}=\frac{2}{3}, \lim _{k \rightarrow \infty} a_{2 k+1}=\frac{1}{3},
	\varliminf _{n \rightarrow \infty} \{a_{n}\}=\frac{1}{3} , \varlimsup _{n \rightarrow \infty} \{a_{n}\}=\frac{2}{3}$

\section{设 $\left\{a_{n}\right\},\left\{b_{n}\right\}$ 是两个有界的正数列, 证明
  $$
	  \begin{array}{l}
		  \varlimsup_{n \rightarrow \infty} a_{n} \cdot \varliminf _{n \rightarrow \infty} b_{n} \leqslant \varlimsup_{n \rightarrow \infty}(a_{n}b_{n})  \leqslant \varlimsup _{n \rightarrow \infty} a_{n} \cdot  \varlimsup_{n \rightarrow \infty} b_{n} \\
		  \varliminf _{n \rightarrow \infty} a_{n} \cdot \varliminf _{n \rightarrow \infty} b_{n} \leqslant \varliminf _{n \rightarrow \infty}\left(a_{n} b_{n}\right) \leqslant \varlimsup _{n \rightarrow \infty} a_{n} \cdot \varliminf _{n \rightarrow \infty} b_{n}
	  \end{array}
  $$}
\textbf{证}\quad
先证第二个不等式右端的不等式.根据定义,存在$ \left\{x_{n}\right\}$ 的一个子数列 $\left\{x_{n_{k}}\right\}$, 使 $x_{n_{k}} \rightarrow \alpha=\varliminf _{n \rightarrow \infty} x_{n} \geqslant 0$ ;

对于数列$y_{n_{k}}$,存在子数列$y_{n_{k_{i}}}$,使$y_{n_{k_{i}}}\rightarrow\beta=\varlimsup_{n \rightarrow \infty}y_{n_{k}}\geqslant 0$.显然 $\varlimsup_{k \rightarrow \infty} y_{n_{k}} \leqslant \varlimsup_{n \rightarrow \infty} y_{n}$. 由于 $x_{n_{k_{i}}} y_{n_{k_{i}}} \rightarrow \alpha \beta$, 故 $\alpha \beta$ 是数列
$\left\{x_{n} y_{n}\right\}$ 的一个聚点.

因此 ,
$\varliminf_{n \rightarrow \infty}\left(x_{n} y_{n}\right) \leqslant \alpha \beta$.由此,再注意到 $\alpha \geqslant 0, \beta \geqslant 0$, 即得
$$
	\varliminf _{n \rightarrow \infty}\left(x_{n} y_{n}\right) \leqslant \alpha \beta \leqslant \alpha\left(\varlimsup_{n \rightarrow \infty} y_{n}\right)=\left(\varliminf_{n \rightarrow \infty} x_{n}\right) \cdot\left(\varlimsup_{n \rightarrow \infty} y_{n}\right)
$$

再证左端的不等式.若 $\varliminf _{n \rightarrow \infty} x_{n}=0$, 则此不等式显然成立,故设 $\varliminf _{n \rightarrow \infty} x_{n}=\beta^{*}>0$ . 于是,存在正整数 $\mathrm{N}_{0}$, 使当
$n>N_{0}$ 时 , $x_{n}>0$ . 根据定义,存在 $\left\{x_{n} y_{n}\right\}$ 的子数列 $\left\{x_{n_{k}} y_{n_{k}}\right\}$, 使
$$
	x_{n_{k}} y_{n_{k}} \rightarrow \alpha^{\prime}=\varliminf_{n \rightarrow \infty}\left(x_{n} y_{n}\right) \geqslant 0
$$

对于数列$ \left\{x_{n_{k}}\right\}$, 存在子数列$ \left\{x_{n_{k_{i}}}\right\}$, 使
$x_{n_{k_{i}}} \rightarrow \beta^{\prime}={\varliminf _{k \rightarrow \infty} x_{n_{k}}}$

注意到 $\beta^{\prime}=\varliminf _{{k \rightarrow \infty}} x_{n_{k}} \geqslant {\varliminf _{n \rightarrow \infty} x_{n}}=\beta^{*}>0$ 以及$x_{n}>0\left(n>N_{0}\right)$, 知
$$
	y_{n_{k_{i}}}=\left(x_{n_{k_{i}}} y_{n_{k_{i}}}\right) \cdot \frac{1}{x_{n_{k_{i}}}} \rightarrow \frac{\alpha^{\prime}}{\beta^{\prime}}
$$

故 $\frac{\alpha^{\prime}}{\beta^{\prime}}$ 是 $\left\{y_{n}\right\}$ 的一个聚点.从而，
$$
	\frac{\alpha^{\prime}}{\beta^{\prime}} \geqslant \varliminf _{n \rightarrow \infty} y_{n}
$$

由此可知
$$
	\varliminf _{n \rightarrow \infty}\left(x_{n} y_{n}\right)=\alpha^{\prime} \geqslant \beta^{\prime}\left(\varliminf _{n \rightarrow \infty} y_{n}\right) \geqslant\left(\varliminf _{n \rightarrow \infty} x_{n}\right) \cdot\left(\varliminf _{n \rightarrow \infty} y_{n}\right)
$$

第一个不等式可类似第二个不等式给出证明.

\section{设 $x_{n}>0, n=1,2, \cdots,$ 且有$\varlimsup_{n \rightarrow \infty} x_{n} \cdot \varlimsup_{n \rightarrow \infty} \frac{1}{x_{n}}=1$.证明:极限 $\lim _{n \rightarrow \infty} x_{n}$ 存在.}
\textbf{证}\quad
$$
	1=\varliminf _{n \rightarrow \infty}\left(x_{n} \cdot \frac{1}{x_{n}}\right) \leqslant\left(\varliminf _{n \rightarrow \infty} x_{n}\right) \cdot\left(\varlimsup_{n \rightarrow \infty} \frac{1}{x_{n}}\right) \leqslant \varlimsup_{n \rightarrow \infty}\left(x_{n} \cdot \frac{1}{x_{n}}\right)=1
$$

故$\left(\varliminf _{n \rightarrow \infty} x_{n}\right) \cdot\left(\varlimsup_{n \rightarrow \infty} \frac{1}{x_{n}}\right)=1$
$$
	\left(\varliminf _{n \rightarrow \infty} x_{n}\right) \cdot\left(\varlimsup_{n \rightarrow \infty} \frac{1}{x_{n}}\right)=\left(\varlimsup_{n \rightarrow \infty} x_{n}\right) \cdot\left(\varlimsup_{n \rightarrow \infty} \frac{1}{x_{n}}\right)
$$

故$\left(\varliminf _{n \rightarrow \infty} x_{n}\right)=\left(\varlimsup_{n \rightarrow \infty} x_{n}\right)$,从而$x_{n}$有极限.

\section{设数列 $\left\{a_{n}\right\},\left\{b_{n}\right\}$ 有界,且满足关系式$a_{n}=b_{n}-q a_{n+1}(0<q<1)$,证明 $:\left\{a_{n}\right\},\left\{b_{n}\right\}$ 有相同的敛散性.}
\textbf{证}\quad
若$b_{n}$收敛，则$\varliminf _{n \rightarrow \infty} a_{n}=\varliminf _{n \rightarrow \infty}\left(b_{n}-q a_{n+1}\right)=\lim _{n \rightarrow \infty} b_{n}-q \cdot \varlimsup _{n-\infty} a_{n+1}$

$\therefore\varliminf _{n \rightarrow \infty} a_{n}+q \cdot \varlimsup _{n\rightarrow\infty} a_{n+1}=\varliminf _{n \rightarrow \infty} a_{n}+q \cdot \varlimsup _{n\rightarrow\infty} a_{n}=\lim _{n \rightarrow \infty} b_{n}$

同理,$\varlimsup _{n \rightarrow \infty} a_{n}+q \cdot \varliminf _{n\rightarrow\infty} a_{n}=\lim _{n \rightarrow \infty} b_{n}$

只有当$\varlimsup _{n \rightarrow \infty} a_{n}=\varliminf _{n \rightarrow \infty} a_{n}$成立时上式才成立,(q不为1即可)所以$a_{n}$收敛

类似的，若$a_{n}$收敛，则可证明$\varliminf _{n \rightarrow \infty} b_{n}=\lim _{n \rightarrow \infty} (1+q)a_{n}=\varlimsup _{n \rightarrow \infty} b_{n}$

故$a_{n}$收敛等价于$b_{n}$收敛,所以两者有相同的敛散性.

\section{设非负数列 $\left\{a_{n}\right\}$ 满足如下的次可加条件:$
	a_{m+n} \leqslant a_{m}+a_{n}$,$  m,n \in \mathbb{N}^{*}$,证明:极限 $\lim _{n \rightarrow \infty} \frac{a_{n}}{n}$ 存在.}
\textbf{证}\quad
由于$x_{n} \leqslant x_{n-1}+x_{1} \leqslant x_{n-2}+x_{1}+x_{1} \leqslant \cdots \leqslant n x_{1}$,任何正整数 $n>N$ 都可表为$n=q N+r$ 的形式,其中 $q$ 为正整数,$r$ 为小于 $N$ 的非负整数 $(0 \leqslant r<N)$.我们有
$$
	x_{n}=x_{q N+r} \leqslant x_{(q-1) N}+x_{N}+x_{r} \leqslant x_{(q-2) N}+x_{N}+x_{N}+x_{r} \leqslant \cdots \leqslant q x_{N}+x_{r} \\\leqslant q x_{N}+r x_{1} \leqslant q x_{N}+N x_{1}
$$

从而,
$$
\frac{x_{n}}{n} \leqslant \frac{q x_{N}}{n}+\frac{N x_{1}}{n} \leqslant \frac{x_{N}}{N}+\frac{N x_{1}}{n}<a+\varepsilon+\frac{N x_{1}}{n}
$$

由此可知 $\varlimsup_{n \rightarrow \infty} \frac{x_{n}}{n} \leqslant a+\varepsilon$, 再根据 $\varepsilon>0 $的任意性,即得 $\varlimsup_{n \rightarrow \infty} \frac{x_{n}}{n} \leqslant a$, 故 $\varliminf _{n \rightarrow \infty} \frac{x_{n}}{n}=\varlimsup_{n \rightarrow \infty} \frac{x_{n}}{n}$

\end{document}